Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 6 - Analytic Trigonometry - Section 6.2 The Inverse Trigonometric Functions (Continued) - 6.2 Assess Your Understanding - Page 480: 36


$\dfrac{ 2 \pi }{3}$

Work Step by Step

We are given: $\cos^{-1} [\sin (\dfrac{7 \pi}{ 6})]$ We know that the trigonometric function $\cos^{-1}$ has domain $[-1,1]$ and range: $[0, \pi]$. Here, $\cos^{-1} (\dfrac{-1}{2})=\dfrac{7\pi}{6}$, because $\cos{(\dfrac{7\pi}{6})}=\dfrac{-1}{2}$ and $\dfrac{7\pi}{6}$ lies in the range of $\cos^{-1}{x}$ . Therefore, we have: $\cos^{-1} [\sin (\dfrac{7 \pi}{ 6})]=\cos^{-1} ({\dfrac{-1}{2}})=\dfrac{ 2 \pi }{3}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.